Follow up on another question I asked recently: Topology: Show restriction of continuous function is continuous, and restriction of a homeomorphism is a homeomorphism
Definition: Let $(X, \mathcal{T})$ and $(Y, \mathcal{J})$ be topological spaces. A function ${\displaystyle f:X\to Y\,}$ is a local homeomorphism if for every point $x \in X$ there exists an open set $U \subseteq X$ containing $x$ and an open set $V \subseteq Y$ such that the restriction ${\displaystyle f|_{U}:U\to V\,}$ is a homeomorphism.
This definition is a bit alarming because it starts with "...if for every point $x \in X$ there exists an open set $U \in \mathcal{T}$...", makes it seem like a property of the underlying space. Can we always find an open $U$? But anyways.
Objective: Show that every local homeomorphism is continuous and open therefore bijective local homeomorphism is a homeomorphism
Proof: (Honestly not sure what I am doing but proceed regardless)
Let $(X, \mathcal{T})$ and $(Y, \mathcal{J})$ be topological spaces and function ${\displaystyle f:X\to Y\,}$ is a local homeomorphism. We will show that $f$ is continuous and open.
First show $f$ is continuous.
$f$ is continuous if for all $V \in \mathcal{J}, f^{-1}(V) \in \mathcal{T}$. Take some $V \in \mathcal{J}$, then $V$ is a subspace equipped with subspace topology $\mathcal{J}_V = \{V \cap W| W \in \mathcal{J}\}$.
Consider the inverse of the restriction $f^{-1}|_U$ on an open set in $\mathcal{J}_V$, then $f^{-1}|_U(V \cap W) = f^{-1}|_U(V) \cap f^{-1}|_U(W) $$= f^{-1}(V) \cap U \cap f^{-1}(W) \cap U = f^{-1}(V) \cap f^{-1}(W) \cap U$.
Then $f^{-1}(V) = f^{-1}(W) \cup U \cup f^{-1}|_U(V \cap W)$. We note all the sets on the right hand side are open. In particular, $U$ is open, $f^{-1}|_U(V \cap W)$ is open by definition of homeomorphism (?? $f^{-1}(W)$ ??), hence $f$ is continuous. ($\leftarrow$ something wrong here!)
Next show $f$ is open.
$f$ is open if $\forall U \in \mathcal{T}, f(U) \in \mathcal{J}$. Consider the restriction $f|_U$ on the subspace topology on $U$, $\mathcal{T}_U = \{U \cap M | M \in \mathcal{T}\}$. $f|_U(U \cap M) = f|_U(U) \cap f|_U(M) = V \cap f(M) \cap f(U)$
Then $f$ is open since $f(U) = f|_U(U \cap M) \cup V \cup f(M)$ and $f|_U(U \cap M)$ is open by definition of homeomorphism, $V$ is open in $\mathcal{T}$ (?? $\cup f(M)$ ??) ($\Leftarrow$ another mistake here)
I'm not quite sure how to proceed with showing bijective + continuous + open + local = homeomorphism part.
Can someone help me fix those two problems and give me some ideas how to conclude that bijective local homeomorphisms are homeomorphisms?